Question

A 1.5-cm-tall object is 90 cm in front of a converging lens that has a 32 cm focal length.

Calculate the image position.

Calculate the image height.

Express your answer to two significant figures and include the appropriate units.

Answer 1

(a) Write the expression of basic lens equation -  

1/p + 1/q = 1/f

where -

p = object distance = +90 cm
q = image distance
f = focal length = + 32 cm

put the values -

1/90 + 1/q = 1/32

1/q = 1/32 - 1/90 = 0.02014

=> q = 49.6 cm = 50.0 cm

Therefore, the image is 50.0 cm behind the converging lens (the sign convention for converging lenses is that positive image distances are behind the lens; positive object distances are in front of the lens)

(b) Again, the magnification is given by m = -q/p =- 49.6 / 90 = -0.55

So, image height = -0.55 x 1.5 = - 0.825 cm = -0.83 cm

The negative sign shows that the image is inverted.